Results 1 to 7 of 7

Math Help - Infinite Limit Proof

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    8

    Infinite Limit Proof

    I need to prove:

    lim (x^4)-(2(x^2)) = +inf
    x->+inf

    Any suggestions?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,

    Hint : x^4-2x^2=x^4\left(1-\frac{2}{x^2}\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    8
    I don't know how to do it. The problem is done and over with, but I still would like to know how to get the answer for my own knowledge.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Calc1stu View Post
    I don't know how to do it. The problem is done and over with, but I still would like to know how to get the answer for my own knowledge.
     \lim_{x\to\infty}x^4\left(1-\frac{2}{x^2}\right)=\lim_{x\to\infty}x^4\cdot\lim  _{x\to\infty}\left(1-\frac{2}{x^2}\right)=\infty\cdot{1}=\infty
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2008
    Posts
    8
    It was supposed to be an epsilon delta proof.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,408
    You want to show that for every M > 0, there exists N > 0 such that whenever x > N, it necessarily follows that f(x) > M  \Leftrightarrow x^4 - 2x^2 > M

    Consider the latter expression as an equality:
    x^4 - 2x^2 = M

    \Leftrightarrow x^4 - 2x^2 - M = 0

    \Leftrightarrow x^2 = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-M)}}{2}

    \Leftrightarrow x^2 = \frac{2 \pm \sqrt{4 + 4M}}{2}

    \Leftrightarrow x^2 = 1 \pm \sqrt{1 + M}

     \Leftrightarrow x^2 = 1 + \sqrt{1 + M} (since x^2 \geq 0 and M > 0)

    \Leftrightarrow x = +\sqrt{1 + \sqrt{1 + M}} (we're considering positive x's)

    So when x \ {\color{red}=} \ \sqrt{1 + \sqrt{1 + M}} we have that f(x)\ {\color{red} =}\ M. We want to show that for some x \ {\color{magenta}>}\ N, we have that f(x) \ {\color{magenta}>}\ M

    Now consider: x \ {\color{magenta}> }\ \sqrt{1 + \sqrt{1 + M}} > 0. Can you finish?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2008
    Posts
    8
    As far as epsilon and delta are concerned which is M and N. I don't know where to go from there. All we ever usually do or did was a basic function.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Infinite Sum Limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 24th 2010, 05:53 PM
  2. Proof of Infinite Limit
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 12th 2009, 10:39 PM
  3. Proof: infinite set minus a fintie set is infinite
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: February 24th 2009, 11:54 PM
  4. Infinite Limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 24th 2008, 07:33 PM
  5. need help plz infinite limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 8th 2006, 08:06 AM

Search Tags


/mathhelpforum @mathhelpforum